How do you find the product of #(t+1)(t^2+2t+4)#?

1 Answer
Aug 13, 2017

Answer:

#(t+1)(t^2+2t+4) = t^3+3t^2+6t+4#

Explanation:

You can use the distributive property to find:

#(t+1)(t^2+2t+4) = t(t^2+2t+4)+1(t^2+2t+4)#

#color(white)((t+1)(t^2+2t+4)) = (t^3+2t^2+4t)+(t^2+2t+4)#

#color(white)((t+1)(t^2+2t+4)) = t^3+2t^2+t^2+4t+2t+4#

#color(white)((t+1)(t^2+2t+4)) = t^3+3t^2+6t+4#

Alternatively, just focus on each power of #t# in descending order and add together the different ways of getting it by multiplying a term in the first factor by a term in the second...

#t^3: " "t * t^2 = t^3#

#t^2: " "t(2t)+1(t^2) = 3t^2#

#t: " "t(4)+1(2t) = 6t#

#1: " "1(4) = 4#

Then add:

#(t+1)(t^2+2t+4) = t^3+3t^2+6t+4#

When looking at each power of #t#, we only really need to multiply the coefficients and add the products, so:

#t^3: " "1 * 1 = 1#

#t^2: " "1 * 2 + 1 * 1 = 3#

#t: " "1 * 4 + 1 * 2 = 6#

#1: " "1 * 4 = 4#

With practice you will probably find that you can normally do these calculations in your head as you write down the product:

#(t+1)(t^2+2t+4) = t^3+3t^2+6t+4#